Let $X$ a Banach space, $f:U\subseteq \mathbb{C}\to X$ an holomorphic function non constant and $N:=\left\{z\in\mathbb{C}:f(z)=0\right\}$. Show that $N$ is closed and discrete.
My idea:
The function $f$ is holomorphic, particulary $f$ is continuos. Then $f^{-1}(\{0\})$ is colsed. Now, suppose that $N$ is not discrete, that is, $N$ has a limit point $\omega\in N$. The function $g=0$ satisfy that $$f(x)=g(x),\hspace{0.2cm}\text{if }x\in N.$$ Finally, $N$ has at least one limit point and cause of the identity principle $$f(x)=g(x)=0\hspace{0.1cm},\text{if }x\in U.$$
Questions:
- Is it right?
- Can I use the hypothesis $f$ non zero instead of non constant?
Thanks.
Identity Principle is for complex valued functions. Here you have to take $x^{*} \in X^{*}$ (where $X^{*}$ is the dual of $X$) and apply Identity Principle for $x^{*} \circ f$ to show that $x^{*} \circ f (z)=0$ for all $z$ for $x^{*}$ (if $N$ has a limit point). This implies $f(z)=0$ for all $z$.